$\text{The table lists the unit selling price of 5 products P, Q, R, S and T. On a particular day, 250 items were sold with average selling price of 60.}$

$\text{The following observations are made:}$

$\text{(i) Quantity of S = 2 × T}$

$\text{(ii) Quantity of R = 3 × T}$

$\text{(iii) Quantity of Q = 4 × T}$

Product	P	Q	R	S	T
Unit S.P.	100	50	40	60	60

$\text{Find the quantity of product P sold on that day.}$

The correct answer is 50.

Let us analyze the given information step-by-step to find the quantity of product P sold on that day.

From the table, the unit selling price (S.P.) for each product is given as follows:
- Product P: 100
- Product Q: 50
- Product R: 40
- Product S: 60
- Product T: 60

Let the quantities of products P, Q, R, S, and T sold on that day be $N_P$, $N_Q$, $N_R$, $N_S$, and $N_T$ respectively.

We are given that the total quantity of items sold is 250:
$N_P+N_Q+N_R+N_S+N_T=250$

We are also given the relationships between the quantities of Q, R, S, and T:
(i) Quantity of S = 2 × T ⇒ $N_S=2N_T$
(ii) Quantity of R = 3 × T ⇒ $N_R=3N_T$
(iii) Quantity of Q = 4 × T ⇒ $N_Q=4N_T$

Substituting these relations into the total quantity equation, we get:
$N_P+4N_T+3N_T+2N_T+N_T=250$
$N_P+10N_T=250$     --- (Equation 1)

Next, we use the information about the average selling price of 60 for the 250 items. The total revenue is:
Total Revenue = Average Selling Price × Total Quantity
Total Revenue = $60\times 250=15000$

We can write the equation for total revenue using the unit prices from the table:
$100N_P+50N_Q+40N_R+60N_S+60N_T=15000$

Substitute $N_Q$, $N_R$, and $N_S$ in terms of $N_T$ into the revenue equation:
$100N_P+50\left(4N_T\right)+40\left(3N_T\right)+60\left(2N_T\right)+60N_T=15000$
$100N_P+200N_T+120N_T+120N_T+60N_T=15000$
$100N_P+500N_T=15000$

Dividing the entire equation by 100 simplifies it to:
$N_P+5N_T=150$     --- (Equation 2)

Now, we have a system of two linear equations:
1) $N_P+10N_T=250$
2) $N_P+5N_T=150$

Subtracting Equation 2 from Equation 1:
$(N_P+10N_T)-(N_P+5N_T)=250-150$
$5N_T=100$
$N_T=20$

Substitute $N_T=20$ back into Equation 2 to find $N_P$:
$N_P+5\left(20\right)=150$
$N_P+100=150$
$N_P=50$

Thus, the quantity of product P sold on that day is 50.